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Chapter 12 Lexicalized and Probabilistic Parsing Guoqiang Shan University of Arizona November 30, 2006
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Outline Probabilistic Context-Free Grammars Probabilistic CYK Parsing PCFG Problems
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Probabilistic Context-Free Grammars Intuition Behind To find “correct” parse for the ambiguous sentences i.e. can you book TWA flights? i.e. the flights include a book Definition of Context-Free Grammar 4-tuple G = (N, Σ, P, S) N: a finite set of non-terminal symbols Σ: a finite set of terminal symbols, where N Λ Σ = Φ P: A β, where A is in N, and β is in (N V Σ)* S: start symbol in N Definition of Probabilistic Context-Free Grammar 5-tuple G = (N, Σ, P, S, D) D: A function P [0,1] to assign a probability to each rule in P Rules are written as A β[p], where p = D(A β) i.e. A a B [0.6], B C D [0.3]
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PCFG Example S NP VP.8 S Aux NP VP.15 S VP.05 NP Det Nom.2 NP ProperN.35 NP Noun.05 NP ProNoun.4 Nom Noun.75 Nom Noun Nom.2 Nom ProperN Nom.05 VP Verb.55 VP Verb NP.4 VP Verb NP NP.05 Det that.5 Det the.8 Det a.15 Noun book.1 Noun flights.5 Noun meal.4 Verb book.3 Verb include.3 Verb want.4 Aux can.4 Aux does.3 Aux do.3 ProperN TWA.4 ProperN Denver.6 Pronoun you.4 Pronoun I.6
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Probability of A Sentence in PCFG Probability of any parse tree T of S P(T,S) = Π D(r(n)) T is the parse tree and S is the sentence to be parsed n is a sub tree of T and r(n) is a rule to expand n Probability of A parse tree P(T,S) = P(T) * P(S|T) A parse tree T uniquely corresponds a sentence S, so P(S|T) = 1 P(T) = P(T,S) Probability of a sentence P(S) = Σ P(T), where T is in τ(S), the set of all the parse trees of S In particular, for an unambiguous sentence, P(S) = P(T)
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Example P(T l ) = 0.15*0.40*0.05* 0.05*0.35*0.75* 0.40*0.40*0.30* 0.40*0.50= 3.78*10 -7 P(T r ) = 0.15*0.40*0.40* 0.05*0.05*0.75* 0.40*0.40*0.30* 0.40*0.50= 4.32*10 -7
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Probabilistic CYK Parsing of PCFG Bottom-Up approach Dynamic Programming: fill the tables of partial solutions to the sub-problems until they contain all the solutions to the entire problem Input CNF: ε-free, each production in form A β or A BC n words, w 1, w 2, …, w n Data Structure Π [i, j, A]: the maximum probability for a constituent with non-terminal A spanning j words from w i β[i, j, A] = {k, B, C}, where A BC, and B spans k words from w i (for rebuilding the parse tree) Output The maximum probability parse will be Π[1,n,1] The root of the parse tree is S, and spans entire string
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Base case Consider the input strings of length one By the rules A w i Recursive case For strings of words of length>1, A → w ij There exists some rules A BC and k 0<k<j B → w ik (known) C → w (i+k)(j-k) (known) Compute the probability of w ij by multiplying the two probabilities If there are more than one A BC, pick the one that maximize the probability of w ij CYK Algorithm Π [i,0,A] {k, B, C} My implementation is in lectura under directory /home/shan/538share/pcyk.c
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PCFG Example – Revisit to rewrite S NP VP.8 S Aux NP VP.15 S VP.05 NP Det Nom.2 NP ProperN.35 NP Noun.05 NP ProNoun.4 Nom Noun.75 Nom Noun Nom.2 Nom ProperN Nom.05 VP Verb.55 VP Verb NP.4 VP Verb NP NP.05 Det that.5 Det the.8 Det a.15 Noun book.1 Noun flights.5 Noun meal.4 Verb book.3 Verb include.3 Verb want.4 Aux can.4 Aux does.3 Aux do.3 ProperN TWA.4 ProperN Denver.6 Pronoun you.4 Pronoun I.6
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Example (CYK Parsing) - Rewrite as CNF S NP VP.8 (S Aux NP VP.15) S Aux NV.15 NV NP VP1.0 (S VP.05) S book.00825 S include.00825 S want.011 S Verb NP.02 S Verb DNP.0025 NP Det Nom.2 (NP ProperN.35) NP TWA.14 NP Denver.21 (NP Nom.05) NP book.00375 NP flights.01875 NP meal.015 NP Noun Nom.01 NP ProperN Nom.0025 (NP ProNoun.4) NP you.16 NP I.24 (Nom Noun.75) Nom book.075 Nom flights.375 Nom meal.3 Nom Noun Nom.2 Nom ProperN Nom.05 (VP Verb.55) VP book.165 VP include.165 VP want.22 VP Verb NP.4 (VP Verb NP NP.05) VP Verb DNP.05 DNP NP NP1.0
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Example (CYK Parsing) – Π matrix Π i+j i 12345 1Aux:.4 2 Pronoun:.4 NP:.16 3 Noun:.1 Verb:.3 VP:.165 Nom:.075 NP:.00375 S:.00825 4 ProperN:.4 NP:.14 5 Noun:.5 Nom:.375 NP:.01875 canyoubookTWAflights
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Example (CYK Parsing) – Π matrix Π i+j i 12345 1Aux:.40 2 Pronoun:.4 NP:.16 S:.02112 NV:.0264 DNP:.0006 3 Noun:.1 Verb:.3 VP:.165 Nom:.075 NP:.00375 S:.00825 S:.00084 VP:.0168 DNP: 000525 4 ProperN:.4 NP:.14 NP:.000375 Nom:.0075 DNP:.002625 5 Noun:.5 Nom:.375 NP:.01875 canyoubookTWAflights
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Example (CYK Parsing) – Π matrix Π i+j i 12345 1Aux:.40S:.01584 2 Pronoun:.4 NP:.16 S:.02112 NV:.0264 DNP:.0006 S:.021504 NV:.002688 3 Noun:.1 Verb:.3 VP:.165 Nom:.075 NP:.00375 S:.00825 S:.00084 VP:.0168 DNP: 000525 S:.00000225 NP:.0000075 Nom:.00015 VP:.000045 DNP:.000001406 4 ProperN:.4 NP:.14 NP:.000375 Nom:.0075 DNP:.002625 5 Noun:.5 Nom:.375 NP:.01875 canyoubookTWAflights
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Example (CYK Parsing) – Π matrix Π i+j i 12345 1Aux:.40S:.01584S:.00016128 2 Pronoun:.4 NP:.16 S:.02112 NV:.0264 DNP:.0006 S:.021504 NV:.002688 S:.00000576 NV:.0000072 DNP:.0000012 3 Noun:.1 Verb:.3 VP:.165 Nom:.075 NP:.00375 S:.00825 S:.00084 VP:.0168 DNP: 000525 S:.00000225 NP:.0000075 Nom:.00015 VP:.000045 DNP:.000001406 4 ProperN:.4 NP:.14 NP:.000375 Nom:.0075 DNP:.002625 5 Noun:.5 Nom:.375 NP:.01875 canyoubookTWAflights
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Example (CYK Parsing) – Π matrix Π i+j i 12345 1Aux:.40S:.01584S:.00016128S:.000000432 2 Pronoun:.4 NP:.16 S:.02112 NV:.0264 DNP:.0006 S:.021504 NV:.002688 S:.00000576 NV:.0000072 DNP:.0000012 3 Noun:.1 Verb:.3 VP:.165 Nom:.075 NP:.00375 S:.00825 S:.00084 VP:.0168 DNP: 000525 S:.00000225 NP:.0000075 Nom:.00015 VP:.000045 DNP:.000001406 4 ProperN:.4 NP:.14 NP:.000375 Nom:.0075 DNP:.002625 5 Noun:.5 Nom:.375 NP:.01875 canyoubookTWAflights
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Example (CYK Parsing) – β matrix B i+j i 12345 1N/A S Aux NV, k = 1 2 S NP VP, k = 1 NV NP VP, k = 1 DNP NP NP, k = 1 S NP VP, k = 1 NV NP VP, k = 1 S NP VP, k = 1 NV NP VP, k = 1 DNP NP NP, k = 1 3 S Verb NP, k = 1 VP Verb NP, k = 1 DNP NP NP, k = 1 S Verb NP, k = 1 NP Noun Nom, k = 1 Nom Noun Nom, k = 1 VP Verb NP, k = 1 DNP NP NP, k = 1 4 NP ProperN Nom, k = 1 Nom ProperN Nom, k = 1 DNP NP NP, k = 1 5 canyoubookTWAflights
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PCFG Problems Independence Assumption Assumption: the expansion of one nonterminal is independent of the expansion of others. However, examination shows that how a node expands is dependent on the location of the node 91% of the subjects are pronouns. She’s able to take her baby to work with her. (91%) Uh, my wife worked until we had a family. (9%) But only 34% of the objects are pronouns. Some laws absolutely prohibit it. (34%) All the people signed confessions. (66%)
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PCFG Problems Lack of sensitivity of words Lexical information in a PCFG can only be represented via the probability of pre-terminal nodes (such as Verb, Noun, Det) However, lexical information and dependencies turns out to be important in modeling syntactic probabilities. Example: Moscow sent more than 100,000 soldiers into Afghanistan. In PCFG, into Afghanistan may attach NP (more than 100,000 soldiers) or VP (sent) Statistics shows that NP attachment is 67% or 52% Thus, PCFG will produce an incorrect result. Why? the word “Send” subcategorizes for a destination, which can be expressed with the preposition “into”. In fact, when the verb is “send”, “into” always attaches to it
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PCFG Problems Coordination ambiguity Look at the following case Example: dogs in houses and cats Semantically, dogs is a better conjunct for cats than houses Thus, the parse [dogs in [ NP houses and cats]] intuitively sounds unnatural, and should be dispreferred. However, PCFG assigns them the same probability, since the structures are using exactly the same rules.
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References NLTK Tutorial: Probabilistic Parsing: http://nltk.sourceforge.net/tutorial/pcfg/index.html http://nltk.sourceforge.net/tutorial/pcfg/index.html Stanford Probabilistic Parsing Group: http://nlp.stanford.edu/projects/stat-parsing.shtml http://nlp.stanford.edu/projects/stat-parsing.shtml General CYK algorithm http://en.wikipedia.org/wiki/CYK_algorithmhttp://en.wikipedia.org/wiki/CYK_algorithm General CYK algorithm web compute http://www2.informatik.hu-berlin.de/~pohl/cyk.php?action=example Probabilistic CYK parsing http://www.ifi.unizh.ch/cl/gschneid/ParserVorl/ParserVorl7.pdf http://catarina.ai.uiuc.edu/ling306/slides/lecture23.pdf
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